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    Hi guys,

    I appreciate that you are all very busy but can anyone help me out with some questions please.

    https://www.undergraduate.study.cam....paper_2016.pdf


    Question 48:

    Im confused by the wording. How can u have two horizontal forces at right angles to each other?

    Actual Answer: A

    Question 53:

    Not a clue. I differentiated and equated to zero to find maxima and minima. Not too sure what to do afterwards.

    Actual Answer: C

    Thanks in advance

    UPDATE: All questions have been answered, thanks @ghostwalker
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    (Original post by Shaanv)
    Hi guys,

    I appreciate that you are all very busy but can anyone help me out with some questions please.

    https://www.undergraduate.study.cam....paper_2016.pdf


    Question 48:

    Im confused by the wording. How can u have two horizontal forces at right angles to each other?

    Actual Answer: A
    One goes North, and one goes East, for example.

    Question 53:

    Not a clue. I differentiated and equated to zero to find maxima and minima. Not too sure what to do afterwards.

    Actual Answer: C

    Thanks in advance
    Do a rough sketch, plotting the turning points - and off to infinities.

    Your k values will equate to horizontal lines at y=k, and need to cut the graph at 4 points.
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    (Original post by ghostwalker)
    Do a rough sketch, plotting the turning points.
    The question says two horizontal forces at right angles to each other. If one is horizontal and the other is perpendicular to it surely the force is vertical.

    Also is there meant to be an attachment
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    (Original post by Shaanv)
    The question says two horizontal forces at right angles to each other. If one is horizontal and the other is perpendicular to it surely the force is vertical.

    Also is there meant to be an attachment
    You only seem to have looked at the last part of my post.

    No attachement on my part.
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    (Original post by ghostwalker)
    You only seem to have looked at the last part of my post.

    No attachement on my part.


    Would this be the correct way to tackle the problem, i dont get an even answer of 5 but it rounds to 5?
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    Shaanv

    Seems to be something wrong with the thread, or my machine, as I can't quote you.

    You've drawn the weight on the same line as one of the forces, which isn't correct.

    As seen from above - see my diagram. The weight will be in the third dimension going into the page/image.

    Work out the resultant of the two forces. This determines the direction of motion. Friction will then oppose that motion - along the same line, but in the opposite direction. Work out the final resultant force, and....

    Name:  Untitled.jpg
Views: 33
Size:  18.8 KB
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    (Original post by ghostwalker)
    Shaanv

    Seems to be something wrong with the thread, or my machine, as I can't quote you.

    You've drawn the weight on the same line as one of the forces, which isn't correct.

    As seen from above - see my diagram. The weight will be in the third dimension going into the page/image.

    Work out the resultant of the two forces. This determines the direction of motion. Friction will then oppose that motion - along the same line, but in the opposite direction. Work out the final resultant force, and....

    Name:  Untitled.jpg
Views: 33
Size:  18.8 KB
    Much appreciated. I only considered a vertical plane which was wrong.

    I wont be making that mistake again.
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